D.S.G. Pollock 283The solution is:
k∗=−(S′∗−^1 S∗)−^1 S∗′−^1 Sk. (6.137)The starting valuesk∗andd∗can be eliminated from the expressions forxandh
in (6.133), which provide the estimates of the components. Thus, using expression
I−P∗=PQfrom (6.118), we get:
h=Sk+S∗k∗
=(I−P∗)Sk=PQSk.(6.138)Then, by using the expression forkfrom (6.132) together with the identityQ′S=
IT, we get:
h=Q(+Q′Q)−^1 Q′y. (6.139)This agrees with (6.128) in the case whereμ=0. The condition thatx+h=y,
which is that the sum of the estimated components equals the data vector, indicates
that:
x=y−h=y−Q(+Q′Q)−^1 Q′y,(6.140)which is equation (6.123) again.
Observe that the filter matrixZη=Q(+Q′Q)−^1 of (6.140), which deliversh=Zηg, differs from the matrixZκ=Q′Zηof (6.132), which deliversk=Zκg,
only in respect of the matrix difference operatorQ′. The effect of omitting the
operator is to remove the need for reinflating the filtered components and thus
to remove the need for the starting values. These matters have been discussed at
greater length by Pollock (2006).
6.8 The Fourier methods of signal extraction
If the data are generated by a stationary stochastic process, then it may be
reasonable to regard them as the product of a circular process, of which the Fourier
representation is readily available. There are some advantages in exploiting the
Fourier representation by performing the essential filtering operations in the fre-
quency domain – for these are usually aimed at suppressing or attenuating some
of the cyclical elements of the data. It is also straightforward to provide a time-
domain interpretation of the frequency domain operations, and the possibility
exists of performing the equivalent operations in either domain.
The dispersion matrix of a circular stochastic process is obtained from the
autocovariance generating functionγ(z)by replacing the argumentzby the cir-
culant matrixKT=[e 1 ,...,eT− 1 ,e 0 ], which is formed from the identity matrix
IT=[e 0 ,e 1 ,...,eT− 1 ]by moving the leading column to the back of the array. In